Let $h(x)=\sec(x)$. Find $h'\left(0\right)$. Choose 1 answer: Choose 1 answer: (Choice A) A $0$ (Choice B) B $2$ (Choice C) C $-1$ (Choice D) D $1$
Solution: Let's first find $h'(x)$. Then, we can evaluate it at $x=0$. Recall that the derivative of $\sec(x)$ is $\dfrac{\sin(x)}{\cos^2(x)}$, or $\sec(x)\tan(x)$. [Is there a way to know this without memorizing?] So $h'(x)=\dfrac{\sin(x)}{\cos^2(x)}$. Now let's plug $x=0$ into $h'$ : $\begin{aligned} &\phantom{=}h'\left({0}\right) \\\\ &=\dfrac{\sin\left({0}\right)}{\cos^2\left({0}\right)} \\\\ &=\dfrac{0}{\left(1\right)^2} \\\\ &=0 \end{aligned}$ In conclusion, $h'\left(0\right)=0$.